Sizing a Motion Axis
Step-by-step motor and drive sizing for horizontal, vertical and rotary axes, with torque, speed, inertia ratio and margin checks.
The sizing method
Sizing a motion axis is a repeatable calculation. The result is a motor and drive whose torque, speed and heat limits comfortably cover the load. The steps are the same for every axis; only the mechanics change.
- 1. Define the move: distance, time, number of moves per cycle, dwell, payload, orientation and external forces.
- 2. Choose the profile and compute peak speed and acceleration (see Motion Control Basics).
- 3. Reflect the load to the motor shaft through the transmission: speed, inertia, and the torque for friction, gravity and process force (see Mechanical Transmission: Screws, Belts and Gearboxes).
- 4. Compute the torque in each phase (accelerate, run, decelerate, dwell) including the motor’s own rotor inertia: T = J × α + friction + gravity.
- 5. Compute the peak torque and speed, and the RMS torque over the whole cycle.
- 6. Choose a motor whose curve contains all the load points with a 20 to 30 percent margin, whose continuous torque exceeds the RMS torque with margin, and with an inertia ratio in range (see Servo Systems and Tuning).
- 7. Size the drive, the supply and any brake or braking resistor; check the voltage, the regeneration and the thermal limits.
- 8. Re-check with the real motor’s inertia, because the first choice changes the answer slightly. Iterate once.
Example A: horizontal ball-screw axis
This is the worked example from Motion Control Basics: 0.5 m in 1.0 s with 0.25 s ramps, a 5 kg load on a 10 mm lead screw. The calculation gave a peak torque of 0.155 N·m at 4 000 rpm and an RMS torque of 0.090 N·m, using a rotor of 2.00 × 10-5 kg·m² (to be confirmed once the motor is chosen).
Worked example: selecting the motor
- With a 30 percent margin: required peak = 0.155 × 1.3 = 0.201 N·m at 4 000 rpm; required continuous = 0.090 × 1.3 = 0.116 N·m.
- Candidate: a 100 W servo with 0.32 N·m continuous, 0.95 N·m peak, 3 000 rpm rated and 5 000 rpm maximum. At 4 000 rpm it is above its rated speed but within the maximum, and the continuous curve is reduced there (about 85 percent at the top speed), still over three times the RMS requirement.
- Inertia ratio: 1.27 × 10-5 ÷ 2.00 × 10-5 = 0.63:1, far below 5:1.
- Peak power is 65 W, a modest fraction of the motor’s 100 W rating plus the drive’s overload.
- Verdict: comfortable. There is room to speed the cycle up, or to select a smaller (50 W) motor and check again.
Example B: a vertical axis
Gravity changes everything. A vertical axis holds a weight all the time (a continuous torque), needs more torque to lift than to lower, and falls if the motor loses torque unless a brake holds it. Take a 20 kg load lifted 300.0 mm in 0.8 s with 0.2 s ramps on a 10 mm ball screw (90 percent efficient).
Worked example: lift and lower
- Profile: v = 0.3 ÷ (0.8 − 0.2) = 0.50 m/s, a = 2.50 m/s². Screw speed 3 000 rpm, angular acceleration 1 571 rad/s².
- Gravity torque: the weight is 196.2 N. Raising needs T = F × lead ÷ (2π η) = 196.2 × 0.010 ÷ (2π × 0.9) = 0.347 N·m all the time. Lowering, the load drives the screw (a ball screw can be back-driven) with 0.281 N·m, which the motor must resist: it works in a braking quadrant.
- Inertia: load 5.07 × 10-5 + rotor and screw 5.00 × 10-5 = 1.01 × 10-4 kg·m²; J × α = 0.158 N·m.
- Torque while lifting: accelerating 0.347 + 0.158 = 0.505 N·m; at constant speed 0.347 N·m; decelerating 0.189 N·m.
- RMS torque over the lift itself = 0.365 N·m, higher than the 0.32 N·m continuous rating of the 100 W servo, so the lifts could not repeat back to back. In a 4 s cycle (0.8 s moving, the rest holding) with a holding brake and zero motor current at rest, the RMS is 0.365 × √(0.8 ÷ 4) = 0.163 N·m: acceptable.
- Without a brake the motor would hold 0.347 N·m all the time, which is above its 0.32 N·m continuous rating: it would overheat. This is why vertical axes need a brake (or a counterbalance), and why gravity often sets the motor size.
- Peak requirement with 30 percent margin: 0.657 N·m at 3 000 rpm, inside the 0.95 N·m peak of the motor. Verdict: the 100 W servo works with a brake and a 4 s cycle; for a faster cycle or no brake, choose the next size up.
Example C: a rotary indexing table
A rotary table of inertia 0.02 kg·m² indexes 90° in 0.5 s, driven through a 10:1 gearbox (90 percent efficient) by a servo with a rotor of 2.00 × 10-5 kg·m², once every 2 s. With no stop in the move a triangular profile (accelerate for half the time, decelerate for the other half) needs the least acceleration.
Worked example: indexing
- Load side: α = 4θ ÷ T² = 4 × 1.57 ÷ 0.25 = 25.1 rad/s²; peak speed ω = 2θ ÷ T = 6.28 rad/s = 60 rpm. Torque at the table T = J × α = 0.02 × 25.1 = 0.503 N·m.
- Motor side: speed 60 × 10 = 600 rpm, acceleration 251 rad/s². The reflected load inertia is 0.02 ÷ 10² = 2.00 × 10-4 kg·m², an inertia ratio of 10.0:1. Without the gearbox the ratio would be 1 000:1.
- Motor torque: to accelerate the rotor Jm × αm = 0.005 N·m, plus the load torque through the gearbox 0.503 ÷ (10 × 0.9) = 0.056 N·m: a peak of 0.061 N·m.
- RMS torque over the 2 s cycle (the table moves for 0.5 s and stops for the rest, with the brake or the servo holding): 0.061 × √(0.5 ÷ 2) = 0.030 N·m. Both the peak and the RMS are well within a small servo; the gearbox lets a motor of a tenth of a newton-meter drive a table needing 0.50 N·m.
Sizing the drive and supply
| Check | Rule | Example A value |
|---|---|---|
| Peak current | Drive peak current ≥ Tpeak ÷ Kt | 0.201 N·m ÷ 0.35 N·m/A = 0.58 A peak |
| Continuous current | Drive continuous current ≥ Trms ÷ Kt | 0.116 ÷ 0.35 = 0.33 A |
| Speed and voltage | Bus voltage above the back-EMF at maximum speed plus losses | Check against the motor’s rated speed and voltage |
| Mechanical power | P = T × ω at the worst point | 65 W |
| Regeneration | Energy to absorb at each stop: ½ J ω² plus the potential energy of a lowered load | ½ × 6.27 × 10-5 × 419² = 5.5 J |
| Brake | Holding torque ≥ 1.5 × the static load torque; fitted on vertical axes | Example B: 0.347 × 1.5 = 0.520 N·m |
Margins and checks
- Torque margin: 20 to 30 percent on the peak and RMS torque for a well-known load; more for friction that is uncertain or varies with temperature and wear.
- Speed margin: the motor’s maximum speed should exceed the peak by at least 10 percent, and the drive’s bus voltage must support it.
- Temperature: catalog ratings assume 25 to 40 °C and a heat-sinked mounting. A motor in a warm enclosure has less continuous torque.
- Inertia ratio: as small as practical (see Servo Systems and Tuning).
- Duty cycle: if the cycle changes (a faster production rate), recompute the RMS.
- Measure afterwards: read the drive’s torque and temperature during real moves, and compare them with the calculation. A big difference means a wrong assumption (friction, mass, efficiency).
The Motor Torque and Inertia Matching Calculator performs the reflection and the ratio check for a gearbox or screw.
Common sizing mistakes
| Mistake | Result | Fix |
|---|---|---|
| Sizing for cruise torque only | The motor cannot accelerate the load | Add J × α and use the peak torque at speed |
| Forgetting the motor’s own inertia | Torque for the acceleration is underestimated | Include the rotor inertia and the screw or gearbox inertia |
| Ignoring gravity on a vertical axis | The motor overheats, or the axis falls when disabled | Include gravity in every phase and fit a brake |
| Taking the catalog torque at the wrong speed | The torque is not available at the working speed | Plot the load points on the speed-torque curve |
| Using the same margin for everything | Over- or under-sizing | Judge how well the loads and friction are known |
| No allowance for efficiency | Gearbox and screw losses are missed | Divide the load torque by the efficiency |
| Not checking regeneration | Overvoltage faults on the drive | Compute the braking energy and add a resistor if needed |
Related
- Motion Control and Actuators: the overview of this guide.
- Motion Control Basics, Choosing a Motor Technology and Mechanical Transmission: Screws, Belts and Gearboxes: the building blocks of the method.
- Braking, Safe Stopping and Troubleshooting: brakes and stopping.
- Motor Torque and Inertia Matching Calculator and Gear Ratio Calculator.
FAQ
How much margin should I add?
Twenty to thirty percent on torque is common for a well-known load. More if friction is uncertain, the temperature is high, or the duty may increase. Test the machine and read the drive’s torque to confirm.
Do I need a brake on a horizontal axis?
Usually not, because the load does not move when the motor is disabled. Add one if a person could be hurt by an unexpected movement or if the axis could coast into something.
Can I use a motor above its rated speed?
Up to the maximum speed the manufacturer gives, with the reduced torque at that speed shown on the curve. Above rated speed the continuous torque falls and the bus voltage must be adequate.
What if the real load is heavier than I thought?
The drive’s torque reading will be higher than predicted: that is the first sign. Re-run the sizing with the measured mass and friction; if the peak or RMS torque is over the limit, reduce the acceleration or step up the motor size.
References
- Kollmorgen, Parker and Siemens motion sizing software and handbooks (the standard calculation).
- Slocum, A. H., Precision Machine Design, SME.
- Pyrhönen, J., Jokinen, T. and Hrabovcová, V., Design of Rotating Electrical Machines, Wiley (thermal limits).
Last reviewed: October 2026. The motor and load values are worked examples; use the manufacturer’s data and a sizing tool for the real axis.